Chapter 3:
Molecular Biology Problems
166
Molecular Biology Problems
If you were a molecular biologist, you would focus on biological molecules
like DNA, RNA, and proteins. Although generally true, your work would overlap
with other areas like genetics and biochemistry. In this chapter, we have given you
problems that will allow you to explore the structure and function of DNA and RNA
and how proteins are elaborated in the cell.
The first problems examine some of the most important experiments that led to the
conclusion that DNA is the genetic material. They are good examples of the history of
science as well as opportunities to analyze real data. Before attempting the question in
this section, refer to your textbook or lecture material.
The first diagnostic question for this chapter is found in section 2. Work through the
diagnostic question on your own and then look at our approach to solving it. If any of
the terms are unfamiliar, consult the appropriate chapter in your textbook.
(1) PROBLEMS EXPLORING CLASSIC EXPERIMENTS
Key words: “Griffith,” “Streptococcus pneumoniae,” “Transforming Principle,” “Avery,”
“Pneumococcus,” “Hershey,” and “Chase.”
Note that this section does not have a diagnostic problem; you should consult your
textbook for further information if you have difficulty working these problems.
(1.1) Frederick Griffith showed that a component of dead bacterial cells could confer
new properties to live bacteria of the same species. The property in question was the
presence of a surface polysaccharide capsule of the bacterium Streptococcus pneumoniae,
the bacterium that causes pneumonia.
Two types of these bacteria are found, based on two main types of surface
polysaccharide:
1) R (rough) cells have no polysaccharide capsule. These bacteria do not cause
disease. They are nonvirulent.
2) S (smooth) cells have a polysaccharide capsule. These bacteria cause disease
and are virulent.
The S (smooth) cells fall into several different subtypes based upon the polysaccharide
capsule. These are designated SI through SXXIII.
You can isolate mutants of all these S strains that no longer make a capsule and are no
longer virulent. The lack of a polysaccharide capsule makes one R strain
indistinguishable from another. However, R strains are also designated R(I) through
R(XXIII), depending upon their origin. An R strain derived from SIII is designated R(III).
Classic DNA Experiments
The central experiment was:
Experiment: Inject a mixture of R(II) and heat-killed SIII into a mouse.
Results: The mouse dies of pneumonia, and virulent SIII bacteria can be isolated from
the mouse.
Griffith et al. also did some control experiments to support experiment 1.
Negative Control: Inject R(II) or heat-killed SIII alone into a mouse.
Results: The mouse lives.
Positive Control: Inject live SIII alone into a mouse.
Results: The mouse dies of pneumonia, and only virulent SIII bacteria can be isolated
from the mouse.
a) For each of the control experiments listed, explain:
•
What alternative models does the result of this control experiment rule out? That
is, complete the sentence, “Without this control result, you could argue that the
mice died in experiment 1 because....”
•
What would it have meant for their model if the results of this control experiment
were the reverse of those expected (the mouse dies instead of living and vice
versa), assuming that the results of experiment 1 were the same as above?
b) Another potential problem with their experiments was the possibility that, at a low
frequency, R(II) can mutate back to the SII from which it was derived. (Note: R(II) cannot
revert to any other subtype of S.) Based on this, why was it essential that they use a
mixture of R(II) and heat-killed SIII instead of a mixture of R(II) and heat-killed SII?
Chapter 3: Molecular Biology Problems
(1.2) The team of Avery, McCarty, and MacLeod was attempting to purify a substance
from smooth (S) pneumococci which was capable of transforming rough (R)
pneumococci into smooth; they called this substance the “transforming substance.”
a) They presumed that the transforming substance was genetic material (a.k.a. genes).
Explain why they believed that this was so.
The authors were trying to distinguish between two models for the “transforming
substance” (genes):
(1) Genes are made of protein.
(2) Genes are made of DNA.
It was their belief that model (2) was correct. They wanted to purify the transforming
substance away from other cellular material and then determine whether it was pure
DNA or a mixture of protein and DNA. They hoped to show that protein was either not
present or not essential for the transforming substance to be able to transform R to S.
b) If they had found traces of protein in their preparations of the transforming substance
(even if it was >99% DNA), this would have made it impossible to rule out model (1).
Explain why this is so.
c) In their paper, they presented an analysis of the elemental composition of several of
their purified preparations. In particular, they compared the ratio of nitrogen to
phosphorus (N/P ratio) of their preparations, with the N/P ratio predicted based on the
structure of DNA.
i) Using the structure of DNA from your text, the atomic weight of N = 15 and the
atomic weight of P = 31, show that the ratio for the DNA:
total mass of N
= 1.69.
total mass of P
€
ii) If their preparations were contaminated with protein, would you expect the
ratio of the preparations to be higher or lower than 1.69? Explain your reasoning.
Classic DNA Experiments
These are their actual data (all of these preparations could transform R to S):
Preparation #
37
38B
42
44
N/P ratio
1.66
1.75
1.69
1.58
iii) What conclusions would you draw from these data, and what would be your
reservations about these conclusions?
d) In another series of experiments, preparations were treated with various enzymes of
known function. They wanted to determine whether these enzymes were capable of
destroying the transforming ability of the preparations. They treated their preparations
with the following enzymes alone or in combination:
• Trypsin: an enzyme that breaks proteins down into amino acids.
• Chymotrypsin: another enzyme that breaks proteins down into amino acids.
None of these treatments had any effect on the transforming ability of their
preparations.
i) Why do these data support model (2)?
ii) Why are these data, on their own, not completely conclusive? If you believed
in model (1), how would you argue that these data are consistent with model (1)?
(1.3) Hershey and Chase provided strong evidence that DNA, and not protein, is the
genetic coding material of the cell through their experiments involving differential
partitioning of 32P-labeled DNA and 35S-labeled protein of bacteriophage T2.
a) In their studies, there were two crucial experiments.
Experiment 1: Bacteriophage were labeled with 32P, which is incorporated only into
DNA.
Experiment 2: Bacteriophage were labeled with 35S, which is incorporated only into
protein.
Chapter 3: Molecular Biology Problems
The bacteriophage were allowed to attach to the bacterial cells and transmit the genetic
material. The bacteria and the bacteriophage were then separated, and the amount of
radioisotope in each was measured. For each scenario below, describe how much
radioactivity you would expect to find in the phage as opposed to the bacteria (for
example: a lot, some of it, little, or none). Explain your reasoning briefly.
i) If the phage contained both DNA and protein but the genetic material injected
was protein and the DNA remains in the phage head.
ii) If the genetic material were a mixture of mostly DNA and a little protein.
iii) If the genetic material were protein that is carried on the DNA and the DNA is
only a scaffold for carrying the protein genetic material.
b) Actually, the data were not as unambiguous as described in most textbooks. (For an
interesting discussion of how this experiment has changed in the telling and retelling,
see “How history has blended,” Nature 249:803-805, 1974.) They performed the
experiment several times and got the following results:
(1) Between 15% and 35% of the32P was found in the supernatant.
(2) Between 18% and 25% of the 35S was found in the pellet.
Assume that their model is correct (the phage injects DNA and not protein).
i) What could have caused the presence of the 32P in the supernatant? How
crucial for their model is it that this number be 0? Why?
ii) What could have caused the presence of some of the 35S in the pellet? How
crucial for their model is it that this number be 0%? Why?
Classic DNA Experiments
c) Based on the above data, which of the models (i, ii, or iii) in part (a) can you rule out?
Explain your reasoning.
(1.4) On a mission to a new solar system, you discover an alien virus that contains
nucleic acids, proteins, and lipids. You also find that this virus can infect E. coli cells,
making it easy to study in the laboratory.
a) You grow this virus with one of the following radioisotopes: 32P, 3H, or 35S.
•
Which of the viral macromolecules will be labeled with 32P?
•
Which of the viral macromolecules will be labeled with 3H?
•
Which of the viral macromolecules will be labeled with 35S?
b) You analyze the nucleic acid and find the following:
Percentage of each base:
A
G
T
C
U
27.6
23.2
28.1
24.0
0.8
What nucleic acid is the virus carrying? How do you know this?
c) Because this is an alien virus, you want to determine which of the macromolecules
(nucleic acids, proteins, and lipids) is the hereditary material. Explain how you would
do this.
Chapter 3: Molecular Biology Problems
d) You examine DNA replication to determine whether it is similar to DNA replication
on Earth. You begin by constructing an in vitro system for DNA replication.
i) Assuming that the process of alien DNA replication is similar to that seen on
Earth, what four components would you include in your system?
ii) Prior to replication, all the template DNA is labeled with 15N. Where is
nitrogen found in DNA?
iii) You repeat the Meselson-Stahl experiments. On the diagram below, draw the
results expected at each round for both conservative and semiconservative
replication.
conservative
14
N-labeled
DNA
15
N-labeled
DNA
N-labeled
DNA
round 1
15
14
round 2
N-labeled
DNA
template
DNA
14
semiconservative
N-labeled
DNA
15
Classic DNA Experiments
N-labeled
DNA
(2) PROBLEMS EXPLORING THE STRUCTURE OF DNA AND RNA
Although biology is often a science of exceptions, there are several important concepts
that apply to the nucleic acids, DNA and RNA. These statements hold in the vast
majority of cases.
Directionality – All strands have a direction. This is specified in terms of the 5! and 3!
carbons in the sugar backbone.
Antiparallel – In order to form proper base pairs, the two strands must run 5! to 3! in
opposite directions.
Base pairing – In DNA, A always pairs only with T. In RNA, A pairs with U. G always
pairs only with C.
Polymerization – New nucleotides are always added only to the 3! end of a chain. Thus,
polymerization always occurs in a 5! to 3! direction.
DNA and RNA Structure:
1) Building blocks of DNA are nucleotides:
Base =
Adenine (A) purines
Guanine (G)
Cytosine (C) pyrimidines
Thymine (T)
O
5'
O P O C H2
O
4'
H
O
H
3'
OH
1'
H
2'
H
Base
H
2)
in DNA
DNA are
are joined
joined by
by 3'
3!––5'5!phosphodiester
phosphodiesterbonds.
bonds.
2) The
The nucleotides
nucleotides in
O
5'
this is an -OH in RNA
C H2
O
P
O
O
Base
5' end
1'
O
4'
H
H
H
H
2'
3'
H
O
5'
Therefore, the
O P O C H2
O
Base
strand has a polarity:
1'
O
4'
H
H
H
H
2'
3'
OH
H
3' end
Chapter 3: Molecular Biology Problems
3) The DNA has two strands held together by hydrogen bonds between the bases on opposite
strands.
hydrogen bonds
CH3
H
O
sugar N
N
O
N
H
H
N
sugar N
N
N
T
H
N
A
N
C
H
N
O
sugar
O
H
N
N
H N
H
N
N
G
sugar
4) The two strands have opposite polarity.
5) The two strands are twisted into a right-handed double-helix.
6) RNA is single-stranded, but it can fold on itself to create regions where hydrogen
bonds form between the bases.
Diagnostic Question:
You are given the sequence of one strand of a DNA molecule:
5! end
A A T C G G C T T A C C T A C C A T T T T A
3! end
a) Directly below the sequence above, give the sequence of the second strand of DNA.
Label all 3! and 5! ends.
b) What chemical group is usually found on the 3! end?
c) What chemical group is usually found on the 5! end?
d) What holds the two strands to each other?
e) A new strand of DNA is made in what direction?
DNA and RNA Structure
Answer to Diagnostic Question:
a)
5! end
Original strand:
A A T C G G C T T A C C T A C C A T T T T A
New strand:
3! end
T T A G C C G A A T G G A T G G T A A A A T
5! end
3! end
b) What chemical group is usually found on the 3! end? There is usually a hydroxyl at the
3! end.
c) What chemical group is usually found on the 5! end? There is usually a phosphate group
at the 5! end.
d) What holds the two strands to each other? Hydrogen bonds hold the two strands together.
e) A new strand of DNA is made in what direction? A new strand of DNA is made in the 5’
to 3’ direction.
Chapter 3: Molecular Biology Problems
(C1) Computer Activity 1. This activity will guide you through the three-dimensional
structure of DNA.
Access “Molecules in 3-d” at this site http://intro.bio.umb.edu/MOOC/jsMol/ and
click on the link for this problem “Molecular Bio C1”.
This software has many useful features, so we will take some time now to describe its
use in detail. This software allows you to get information from the image in several
ways:
• Rotating the molecule: This is the best way to get an idea of the molecule’s threedimensional structure. You can click and drag on any part of the molecule and it
will rotate as though you had grabbed it.
• Zooming in or out: This helps to get close-up or “big-picture” views of the
molecule. Hold the shift key down while dragging the cursor up (to zoom out) or
down (to zoom in) the image.
• Identifying the atom you are looking at: You can find information on the atoms in
the molecule in one of two ways:
o By clicking on an atom and looking at the lower left of the “Molecules in 3dimensions” window. A small line of text will appear there with
information on the atom you just clicked.
o By putting the cursor over the atom you are interested in and waiting a
few seconds for the information to pop up. The program will then display
information on the atom in a little pop-up window. The information in the
pop-up window is more detailed than the above but rather cryptic. Try
putting the cursor over some other atoms to see what you get. Note that
this does not always work, especially on Macintosh computers.
In addition to the above, atoms are also identified by their color. The color scheme is
shown to the right of the molecule images.
Here are a few important notes about the way these molecules are displayed in this
problem:
• The atoms are colored according to the scheme at the right of the window.
• Only covalent bonds are shown.
• All covalent bonds are shown as single lines; single, double, and triple bonds
are all shown identically. You have to figure out the bond type based on your
knowledge of covalent bonding and DNA structure.
• Hydrogen atoms are not shown in these structures. This is because these are
actual protein structures determined by X-ray crystallography. Hydrogen
atoms are not visible in X-ray crystallograms and are therefore not shown in
these structures. You have to infer the hydrogens yourself based on your
knowledge of covalent bonds and DNA structures.
DNA and RNA Structure
a) Click the button marked “Load First DNA Molecule.” You will see a black window
with a DNA molecule shown in “spacefill” mode where atoms are shown as solid
spheres at their actual sizes. You can click on the ”Show atoms as ball and stick” button
to change the representation to “ball and stick” where atoms are shown as balls and the
covalent bonds connecting them are shown as rods. You will find it useful to switch
back and forth between the two views. Note that you may sometimes need to click this
button three times to get the view to change.
• The ball and stick view shows covalent bonds as rods and is most useful for
determining which atoms are covalently bonded to each other.
• The spacefill view shows atoms as joined spheres of their approximate actual size
in the molecule. It is most useful for determining which atoms are closest together.
Because it does not show covalent bonds, it can sometimes be hard to figure out
which atoms are covalently bonded.
i) Click on the “Show the two strands” button and select the “spacefill” view. In
this view, the atoms are colored as follows:
• One sugar-phosphate backbone strand is pink.
• The other sugar-phosphate backbone strand is light yellow.
• The 5’ carbons at the ends of the two backbone strands are purple.
• The 3’ carbons at the ends of the two backbone strands are white.
• The bases are green.
The two sugar-phosphate backbone strands run next to each other. Based on the
structure, are they parallel (both run 5’ to 3’ in the same direction) or antiparallel
(run 5’ to 3’ in opposite directions)?
ii) Switch to a “ball and stick” view. Based on this view, are there any covalent
bonds between the bases on the two different strands?
iii) Click on the “Color-code bases and the two strands” button. This is the same
as (i) and (ii), except the bases are colored to identify which type they are.
• Adenine is yellow.
• Thymine is red.
• Cytosine is blue.
• Guanine is green.
Based on this view, which bases pair with which? By clicking on individual
atoms, you can identify the different bases as you click along the chain. Using
this technique, determine the sequence of as much of both DNA strands as you
can.
Chapter 3: Molecular Biology Problems
b) Click the “Load Second DNA Molecule” button. You will see a short (3-base-pair)
double-stranded DNA. This illustrates the details of DNA structure. You should switch
to the “ball and stick” view to help you see this more clearly. The atoms are colored
using the color scheme on the right of the window except:
• 5’ carbons are purple.
• 3’ carbons are white.
i) Look at the 5’ and 3’ carbons and determine the 5’ to 3’ direction of each strand.
Note that every nucleotide contains both a 5’ and a 3’ carbon. If you look at the
strands, you will see that, in one strand, the 5’ carbons in each sugar come first,
while in the other it is reversed.
ii) Determine the sequence of both strands of the short DNA molecule shown.
The “spacefill” view may be best for this. Hint: you don’t have to look at every
atom in the bases: the bases are colored as described. Also, you can click on
atoms in each base and they will be identified in the information line at the
bottom of the window.
DNA and RNA Structure
(2.2) Consider the following DNA segment that has begun to unwind for replication.
The arrow represents the first nucleotides of the newly formed DNA. The arrow point
marks the site where a new nucleotide will be added.
a) What is the sequence of the newly formed DNA?
b) Label all 5! and 3! ends.
c) Draw an arrow on the bottom strand such that the arrow point marks the site where
a new nucleotide will be added.
(2.3) Look at the following schematics of RNA molecules. The hydrogen bonds
between the bases are indicated by the lines.
a) Given that base pairing requires the strands to be antiparallel, circle the RNA
molecules that could form.
3!
3!
5!
5!
5!
3!
3!
3!
3!
5!
5!
5!
b) If the sequence in a region were 5!… ACGGACGC…3!, what would be the sequence
of the DNA that is hydrogen bonded to it?
Chapter 3: Molecular Biology Problems
(2.4) The next nucleotide to be added to a growing DNA strand is dCTP (shown).
• Circle the part of the growing DNA chain to which the next base is attached.
• Circle the part of the dCTP that is incorporated into the growing DNA chain.
Growing DNA
chain
C 2 O
HC
C
O
H
C 3
O
H
C
C
H
N
O
-O
N
O
DNA and RNA Structure
P
-O
O
O
P
-O
O
O
dCT
P
P
O
-O
CH 2
NH 2
N
O
C
C
C
C
OH
H
N
O
(3) DNA REPLICATION
Diagnostic Question:
Shown below are two sequential close-up views of the same replication fork.
DNA
template
DNA synthesized
time =
before
1
DNA synthesized
time = 1 and time =
between
2
time =
1
time =
2
Arrows indicate the direction of DNA
polymerization.
a) Label the 5! and 3! ends of all the DNA molecules shown.
b) The sequence within the circled region on the top template strand was as follows:
5! … ATTCCG …3!
•
Give the sequence of the complementary DNA.
•
Box the area where this complementary sequence is found.
c) As the two template strands are pulled apart, synthesis of new DNA using the bottom
template strand is continuous. Why is synthesis of new DNA using the top strand
discontinuous?
d) Label the leading strand and the lagging strand.
Chapter 3: Molecular Biology Problems
Answer to Diagnostic Question:
DNA
template
DNA synthesized
time =
before
1
DNA synthesized
time = 1 and time =
between
2
5!
3!
5!
5!
Lagging
3!
3!
5!
3!
5!
3!
3!
5!
time =
1
3!
5!
3!
Leading
5!
3!
5!
time =
2
Arrows indicate the direction of DNA
polymerization.
The answer to this question is really a restatement of some of the concepts given at the
beginning of this section.
• Directionality – All DNA strands have a direction. This is specified in terms of the
5! and 3! carbons in the sugar backbone.
• Antiparallel – In order to form proper base pairs, the two strands must run 5! to 3! in
opposite directions.
• Base pairing – A always pairs only with T. G always pairs only with C.
• Polymerization – New nucleotides are always added only to the 3’ end of a chain.
Thus, polymerization always occurs in a 5’ to 3’ direction.
a) See above.
b) The sequence of the complementary DNA would be: 3!…TAAGGC…5!. This
sequence is boxed in two places on the diagram above.
c) DNA replication requires a template, and new nucleotides are added only to the 3!
end of a chain. As the two original strands are pulled apart, a template becomes
available and is copied such that the new DNA is made in the 5! to 3! direction. At
time = 2, an additional template is available and is copied such that the new DNA is
made in the 5! to 3! direction. The directional constraints of DNA replication cause the
top template strand to be copied in a discontinuous fashion.
d) See above.
DNA Replication
(3.1) DNA replication involves many different enzymatic activities. Match each enzyme
activity listed below with the function(s) that it has in the replication process. The first
one is done for you.
Enzyme Activity
Topoisomerase
Function
k
Primase (synthesizes primer)
DNA polymerase to elongate
new DNA strand
Helicase to unwind DNA
DNA polymerase to replace
RNA with DNA
Processivity factor
Choose from:
a) 3!⇒ 5! growth of new
DNA strand
b) 5!⇒ 3! growth of new
DNA strand
c) 3!⇒ 5! exonuclease
d) 5!⇒ 3! exonuclease
e) Makes RNA primer
complementary to the
lagging strand
f) Makes RNA primer
complementary to the
leading strand
g) Makes peptide bonds
h) Separates the two DNA
strands
i) Maintains DNA
polymerase on template
j) Provides 3!-hydroxyl for
initiation of DNA
polymerization
k) Untangles super-coiled
DNA
(3.2) The following diagram shows a replication bubble within a DNA strand. A primer
for DNA replication is 5’-GUACGUUG-3’.
a) Draw the primer where it would anneal in the replication bubble and indicate the
direction of replication by an arrow.
5'5!
G T C C G A C
3'3!
C A G G C T G
Chapter 3: Molecular Biology Problems
A T T G G C C
T A A C C G G
b) If the replication fork moves to the left, will this primer be used to create the leading
strand or the lagging strand? Please explain your answer.
c) If you answered leading strand, explain why replication is continuous. However, if
you answered lagging strand, explain why replication is discontinuous.
(3.3) Shown is a representation of an origin of replication. Synthesis of new DNA occurs
on both strands and in both directions.
template 3
template 1
CAAGG
GGAAC
A
C
3!
5!
5!
3!
D
B
template 2
GGAAC template 4
CAAGG
fork 1
ORI
fork 2
a) For the following, use sites A and B with respect to fork 1 and sites C and D with
respect to fork 2.
i) On which strand(s) will replication be continuous?
template 1
template 2
template 3
template 4
ii) To which site or sites (A, B, C, or D) can the primer 5!-GUUCC-3! bind to
initiate replication?
DNA Replication
iii) When DNA ligase is inhibited, it differentially affects the synthesis from the
leading and the lagging strands. Explain which strand (leading or lagging) is
more affected by the lack of DNA ligase and why.
(3.4) From this origin, replication is occurring on both strands and the two forks are
moving away from each other.
origin of
replication
a) Label the 3! and 5! ends of the five DNA strands shown. Indicate which strands are
Okazaki fragments.
b) What enzyme is required at C in the diagram above?
c) To which site (A or B or both) can the primer, 5!-CAAGG-3! bind to initiate
replication?
d) For each site chosen (in iii), what is the direction of elongation (left or right) of the
daughter DNA strand?
e) For each site chosen (in iii), is DNA synthesis performed in a continuous or a
discontinuous fashion relative to the nearest replication fork?
Chapter 3: Molecular Biology Problems
(3.5) Imagine that the DNA sequence adjacent to position 90 functions as an origin of
replication.
base pair 90
base pair 1
The sequence in these regions is:
80
100
#
#
5!…G C A T G C G T A C A A T A G T T C G A C …3!
3!…C G T A C G C A T G T T A T C A A G C T G …5!
a) What is the sequence of the RNA primer that binds to the top strand at base pair
positions 80-90? Indicate the 5! and 3! ends of your primer.
b) Would DNA synthesis from the primer in (a) above be continuous or discontinuous?
c) What is the sequence of the RNA primer that binds to the bottom strand at base pair
positions 90-100? Indicate the 5! and 3! ends of your primer.
d) Would DNA synthesis from the primer in (c) above be continuous or discontinuous?
DNA Replication
(4) TRANSCRIPTION AND TRANSLATION
(4.1) Transcription and translation in prokaryotes
Some important notes:
• The promoter dictates where transcription starts. The promoter specifies the
beginning point and the direction of transcription.
• The promoter is a site on the double-stranded DNA molecule where RNA
polymerase binds.
• The concepts governing DNA synthesis also apply to RNA synthesis. Review
directionality, base pairing, and polymerization.
• Translation starts with the AUG closest to the 5! end of the mRNA.* This AUG
need not be at the 5! end of the mRNA, nor does it have to be a multiple of three
nucleotides from the 5! end of the mRNA.
• Translation ends with the first in-frame stop codon, even if more nucleotides
remain in the mRNA.
• Translation can restart at the next start codon following a stop codon.
* Please note that this is an oversimplification used in the context of these problems. In
real genes, the start codon must be preceded by a particular sequence in order to be
recognized as a start codon.
Chapter 3: Molecular Biology Problems
Here are two different ways to show the genetic code:
U
C
A
G
U
UUU phe (F)
UUC phe (F)
UUA leu (L)
UUG leu (L)
CUU leu (L)
CUC leu (L)
CUA leu (L)
CUG leu (L)
AUU ile (I)
AUC ile (I)
AUA ile (I)
AUG met (M)
GUU val (V)
GUC val (V)
GUA val (V)
GUG val (V)
UCU
UCC
UCA
UCG
CCU
CCC
CCA
CCG
ACU
ACC
ACA
ACG
GCU
GCC
GCA
GCG
C
ser (S)
ser (S)
ser (S)
ser (S)
pro (P)
pro (P)
pro (P)
pro (P)
thr (T)
thr (T)
thr (T)
thr (T)
ala (A)
ala (A)
ala (A)
ala (A)
A
UAU tyr (Y)
UAC tyr (Y)
UAA STOP
UAG STOP
CAU his (H)
CAC his (H)
CAA gln (Q)
CAG gln (Q)
AAU asn (N)
AAC asn (N)
AAA lys (K)
AAG lys (K)
GAU asp (D)
GAC asp (D)
GAA glu (E)
GAG glu (E)
Transcription and Translation
G
UGU cys (C)
UGC cys (C)
UGA STOP
UGG trp (W)
CGU arg (R)
CGC arg (R)
CGA arg (R)
CGG arg (R)
AGU ser (S)
AGC ser (S)
AGA arg (R)
AGG arg (R)
GGU gly (G)
GGC gly (G)
GGA gly (G)
GGG gly (G)
U
C
A
G
U
C
A
G
U
C
A
G
U
C
A
G
Diagnostic Question:
The statement “DNA goes to RNA goes to protein” is used as a shorthand description of
the flow of information in the cell.
a) “DNA goes to RNA” is meant to describe the process of _______________________.
For this process to occur, we need an enzyme called _____________________ that uses
DNA as a template to make _________________________.
b) “RNA goes to protein” is meant to describe the process of _______________________.
Three different types of RNA are used in this process. _________________ RNA is used
as the template to make a _____________________. ___________RNA is part of the
complex called a ______________ required for protein synthesis. __________RNA is a
small RNA molecule that acts as an adapter between RNA and protein.
c) Shown below is a schematic of a prokaryotic gene. The stippled area represents the
region that is transcribed.
The
arrow.
represents the promoter, and the direction of transcription is shown by the
5!
3!
a) Which strand of the DNA is being used as a template for transcription, the top or the
bottom? Why?
b) On the diagram, indicate where the start codon would be. What sequence would it
have?
c) On the diagram, indicate where the stop codon would be.
d) On the diagram, indicate where the transcription stop would be found.
Chapter 3: Molecular Biology Problems
Answer to Diagnostic Question:
The statement “DNA goes to RNA goes to protein” is used as a shorthand description of
the flow of information in the cell.
a) “DNA goes to RNA” is meant to describe the process of Transcription. For this
process to occur, we need an enzyme called RNA polymerase that uses DNA as a
template to make messenger RNA.
b) “RNA goes to protein” is meant to describe the process of Translation. Three
different types of RNA are used in this process. Messenger RNA is used as the
template to make a polypeptide (protein). Ribosomal RNA is part of the complex called
a ribosome required for protein synthesis. Transfer RNA is a small RNA molecule that
acts as an adapter between RNA and protein.
c) Shown below is a schematic of a prokaryotic gene. The stippled area represents the
region that is transcribed.
The
arrow.
represents the promoter, and the direction of transcription is shown by the
Transcription stop
5!
3!
Start Codon
Stop Codon
a) Which strand of the DNA is being used as a template for transcription, the top or the
bottom?
The bottom strand is used as a template for transcription. RNA polymerase binds at the
promoter and moves in the direction of the arrow. RNA can be made only in the 5! to 3! direction,
antiparallel and complementary to the template. Thus, the template must be the bottom strand.
b) On the diagram, indicate where the start codon would be. What sequence would it
have?
The exact position cannot be determined. In general, the start codon will be near the promoter
and the sequence will be AUG.
c) On the diagram, indicate where the stop codon would be.
The exact position cannot be determined. In general, the stop codon will be near the transcription
stop and could have the sequence UAA, UAG, UGA.
d) On the diagram, indicate where the transcription stop would be found.
Because you were told that the stippled area is the region transcribed, the exact position can be
determined.
Transcription and Translation
(4.1.1) Shown below are three genes (gene 1, gene 2, and gene 3) located on the same
bacterial chromosome.
a) Indicate where on the diagram you would find the following for each gene:
• Promoter (p1 for gene 1, p2 for gene 2, and p3 for gene 3)
• Transcription termination site (tts1, tts2, and tts3)
• Start codon (start1, start2, and start3)
• Stop codon (stop1, stop2, and stop3)
• Template strand (ts1, ts2, and ts3), the DNA strand that directs RNA synthesis
Be sure to indicate the component on the appropriate molecule (DNA or RNA). The
origin of replication is shown as a sample: the replication complex recognizes the origin
of replication as a double-stranded DNA sequence (not an RNA sequence), so the origin
(ori) is shown as a small region of the DNA.
5!
DNA 5'
3'
3!
ori
Gene 1
Gene 3
Gene 2
3!
3'
5'
5!
transcription
5!5'
3!
3'
mRNA 1
5'
5!
3'
3!
mRNA 2
5!
5'
3'
3!
mRNA 3
translation
protein 1
protein 2
protein 3
b) If a mutation inactivates the promoter for gene 2, do you expect protein 2 to be
produced? Briefly explain.
c) After mutating the promoter for gene 2, you discover that protein 2 is still being
made. Further analysis reveals that a spontaneous mutation has occurred in the
transcription termination site in gene 1. How can you explain the presence of protein 2?
Chapter 3: Molecular Biology Problems
(4.1.2) Shown below is an 80 base pair segment of a hypothetical gene. It includes the
promoter and the first codons of the gene. The sequences of both strands of the DNA
duplex are shown: the top strand reads 5! to 3! left to right (1 to 80); the bottom,
complimentary, strand reads 5! to 3! right to left (80 to 1).
a) Synthesis of the mRNA starts at the boxed A/T base pair indicated by the (a) below
(#11) and proceeds left to right on the sequence below. Write the sequence of the first 10
nucleotides of the resulting mRNA.
5!-TGTTGTGTGGAATTGTGAATGGATAACAATGTGACACAGGAAACAGCTAAGACCATGTTTTGACCAAGCTCGGAATTAAC-3!
1 ---------+a--------+---------+---------+---------+---------+---------+---------+ 80
3!-ACAACACACCTTAACACTTACCTATTGTTACACTGTGTCCTTTGTCGATTCTGGTACAAAACTGGTTCGAGCCTTAATTG-5!
b) Suppose the synthesis of mRNA started at the boxed T/A base pair indicated by the
(b) below (#77), and proceeded right to left. What would be the first five nucleotides of
the mRNA?
5!-TGTTGTGTGGAATTGTGAATGGATAACAATGTGACACAGGAAACAGCTAAGACCATGTTTTGACCAAGCTCGGAATTAAC-3!
1 ---------+---------+---------+---------+---------+---------+---------+------b--+ 80
3!-ACAACACACCTTAACACTTACCTATTGTTACACTGTGTCCTTTGTCGATTCTGGTACAAAACTGGTTCGAGCCTTAATTG-5!
c) The mRNA you just wrote has almost the same sequence as one of the DNA strands.
Which DNA strand is this? What is the difference between it and the mRNA sequence?
d) What are the first three amino acids of the polypeptide that would result from
translation of the mRNA from part (a)? A table of the genetic code can be found in your
textbook.
e) Does translation terminate at the UAA in the mRNA corresponding to the boxed
bases at positions 48-50? Why or why not?
Transcription and Translation
f) What are the last three amino acids of the polypeptide that would result from
translation of the mRNA from part (a)?
g) Mutations can add base pairs, delete base pairs, or change base pairs. The sequence
shown below is the same as in part (a), except a G/C base pair has been added between
30 and 31. Note that the overall length of the DNA is now 81 base pairs due to the
addition. This is a frame-shift mutation.
5!-TGTTGTGTGGAATTGTGAATGGATAACAATGGTGACACAGGAAACAGCTAAGACCATGTTTTGACCAAGCTCGGAATTAAC-3!
1 ---------+---------+---------+---------+---------+---------+---------+---------+3!-ACAACACACCTTAACACTTACCTATTGTTACCACTGTGTCCTTTGTCGATTCTGGTACAAAACTGGTTCGAGCCTTAATTG-5!
What will the sequence of the resulting polypeptide be?
h) The sequence shown below is the same as in part (a), except the C/G base pair at 27
has been deleted. Note that the DNA is one base pair shorter. This is a frame-shift
mutation.
C/G deleted from here
5!-TGTTGTGTGGAATTGTGAATGGATAAAATGTGACACAGGAAACAGCTAAGACCATGTTTTGACCAAGCTCGGAATTAAC-3!
1 ---------+---------+--------+---------+---------+---------+---------+--------- 79
3!-ACAACACACCTTAACACTTACCTATTTTACACTGTGTCCTTTGTCGATTCTGGTACAAAACTGGTTCGAGCCTTAATTG-5!
What will the sequence of the resulting protein be?
Chapter 3: Molecular Biology Problems
i) The sequence shown below is the same as in part (a), except the T/A base pair at 30
has been changed to a G/C base pair.
5!-TGTTGTGTGGAATTGTGAATGGATAACAAGGTGACACAGGAAACAGCTAAGACCATGTTTTGACCAAGCTCGGAATTAAC-3!
1 ---------+---------+---------+---------+---------+---------+---------+---------+ 80
3!-ACAACACACCTTAACACTTACCTATTGTTCCACTGTGTCCTTTGTCGATTCTGGTACAAAACTGGTTCGAGCCTTAATTG-5!
What will the sequence of the resulting protein be? Is this a silent mutation (no change
in amino acid sequence), missense mutation (one amino acid changed to a different
amino acid), or nonsense mutation (one amino acid changed to a stop codon)?
(4.1.3) The sequences of both strands of a DNA duplex are shown below. The top
strand reads 5! to 3! left to right (1 to 120) and the bottom, complementary, strand reads
5! to 3! right to left (120 to 1). The letters above or below the underlined nucleotides are
keyed to the particular questions that follow.
a
f
h gi
5!-GACCACACCAGGCCCACTAGACTAGGTAATTTCACACAGGAAACAGCTATGGCCATGAGC
1 ---------+---------+---------+---------+---------+---------+60
3!-CTGGTGTGGTCCGGGTGATCTGATCCATTAAAGTGTGTCCTTTGTCGATACCGGTACTCG
a
f
h gi
d
ACGCCAAGCTCGGAATTAACCCTCATTAAAGGGAACCGAGGCTGAAGCTCCACCGCGGTG-3!
61 ---------+---------+---------+---------+---------+--------- 120
TGCGGTTCGAGCCTTAATTGGGAGTAATTTCCCTTGGCTCCGACTTCGAGGTGGCGCCAC-5!
c
a) Assume that transcription (synthesis of mRNA) begins at the underlined A/T
(written as top strand/bottom strand) at base pair 41 and proceeds to the right as shown
on the diagram. What are the first 15 nucleotides of the resulting mRNA synthesized?
b) Given the mRNA synthesized in part (a), what are the first six amino acids of the
resulting protein?
Transcription and Translation
c) Given the mRNA synthesized in part (a), does protein translation end with the
underlined TAA sequence on the bottom strand at nucleotides 85, 86, and 87? Why or
why not?
d) Given the mRNA synthesized in part (a), does protein translation end with the
underlined TAA sequence on the top strand at nucleotides 87, 88, and 89? Why or why
not?
e) Given the mRNA synthesized in part (a), what would be the last three amino acids of
the resulting protein?
f) Given the mRNA synthesized in part (a), what would the changes to the sequence of
the resulting protein be if the underlined A/T base pair (written as top strand/bottom
strand) at position 45 was deleted from the DNA sequence by mutation? Explain
briefly.
g) Given the mRNA in part (a), what would the changes to the sequence of the resulting
protein be if the underlined C/G base pair (written as top strand/bottom strand) at
position 54 was changed to an A/T base pair? Explain briefly.
h) Given the mRNA synthesized in part (a), what would the changes to the sequence of
the resulting protein be if the underlined G/C base pair (written as top strand/bottom
strand) at position 52 was changed to a C/G base pair? Explain briefly.
Chapter 3: Molecular Biology Problems
i) Given the mRNA in part (a), what would the changes to the sequence of the resulting
protein be if the underlined A/T base pair (written as top strand/bottom strand) at
position 55 was deleted by mutation? Explain briefly.
(4.1.4) You are studying a protein and wish to determine the nucleotide sequence that
encodes its first four amino acids. The sequence of the first four amino acids of the
normal protein is as follows:
N – Met – Ser – Cys – Trp – ......... – C
a) What possible mRNA sequences could encode the first four amino acids shown
above? You need not write out all combinations; just write out all the codons that could
encode each amino acid in their proper order.
HINT: there are six codons for serine (Ser). Be sure to indicate the 5! and 3! ends as
appropriate.
Transcription and Translation
b) You generate a mutant that produces an altered version of this protein due to a single
base substitution in codon 2. The resulting protein sequence (differences from normal
are bold underlined) is:
N – Met – Gly – Cys – Trp – ......... – C
What does this tell you about the nucleotide sequence of the mRNA that encodes the
serine (Ser) in the normal protein shown at the top of this page?
c) You generate a different mutant that produces an altered version of the protein shown
at the top of this page. This mutation is due to a single base deletion of the first
nucleotide of codon 2. The resulting protein sequence is shown below (differences from
normal are bold underlined). Note that the mutation from part (b) is not present in this
mutant.
N – Met – Val – Ala – Gly – ......... – C
Based on this information, what is the nucleotide sequence of the mRNA that encodes
the first four amino acids of the normal protein?
(4.1.5) You are studying a bacterium-like organism discovered on the wreck of an alien
spacecraft. You discover that the creature has DNA and RNA like Earth organisms,
except they have four different bases, which you name W, X, Y, and Z to differentiate
them from the familiar A, T, G, and C. You also discover that the creature has
transcription and translation processes almost identical to Earth cells – genes are
double-stranded DNA, transcribed to mRNA, which is translated using tRNA and
ribosomes.
You do notice one unusual feature: this creature’s proteins are composed of only 14
amino acids, instead of the usual 20. You know that the organism’s codons contain no
more than three nucleotides. The amino acids used are as follows:
arginine
isoleucine
phenylalanine
tryptophan
aspartic acid
leucine
proline
valine
glutamic acid
lysine
serine
Chapter 3: Molecular Biology Problems
glutamine
methionine
threonine
You decide to decipher the genetic code on this organism. First, you find conditions
where you can react synthetic mRNAs with alien ribosomes, tRNA, and all the
components needed to promote translation. You find that, under these unusual in vitro
conditions, translation initiation can occur without a start codon and translation
termination can occur without a stop codon. This means that initiation and termination
can randomly occur on the mRNAs, leading to short oligopeptides.
First, you must find out how many nucleotides per codon there are.
For your first round of experiments, you add synthetic RNA to an alien translation
reaction mix and sequence the protein that is synthesized. The results are as follows:
Synthetic RNA added
W-W-W...(W)n
(X)n
(Y)n
(Z)n
Protein produced
Met-Met-Met...(Met)m
(Val)m
(Thr)m
(Leu)m
a) Can you conclude anything about the number of nucleotides per codon from this
information?
You perform a second round of experiments using the same procedure:
Synthetic RNA added
(WX)n
(WY)n
(WZ)n
(XY)n
(XZ)n
(YZ)n
Protein produced
a mixture of (Ile)m and (Glu)m
(Lys)m
a mixture of (Arg)m and (Phe)m
a mixture of (Asp)m and (Gln)m
a mixture of (Pro)m and (Ser)m
(Trp)m
b) From this information you are able to deduce the number of nucleotides per codon.
What is this number and how did you deduce it?
Transcription and Translation
You perform a final round of similar experiments:
Synthetic RNA added
(WXY)n
Protein produced
(Glu-Lys-Gln)m
[note: this is equivalent to (Gln-Glu-Lys)m
and (Lys-Gln-Glu)m]
(WWX)n
(Ile-Glu-Met)m [same note]
(XYZ)n
(Ser-Trp-Gln)m [same note]
(WXZ)n
(Pro-Glu-Arg)m [same note]
(WZY)n
Phe-Lys
(XWY)n
Ile-Asp
Note that the last two are short peptides of only two amino acids.
c) From all the experiments above, you are able to deduce the genetic code in this
organism. What is it? List all the codons and the amino acids they encode. Include stop
codons.
Chapter 3: Molecular Biology Problems
(4.1.6) The following transcription and translation graphic is incomplete.
a) Finish the diagram by completing i to v.
ribosome
mRNA
RNA Polymerase
5!5’
3!3’
GGGCATGCGCCCTACGTAAACGGCATAAAGCTCCCCAGTTTGCGCGCGCATTGTGATG
Promoter
CCCGTACGCGGGATGCATTTGCCGTATTTCGAGGGGTCAAACGCGCGCGTAACACTAC
i) Label 5! and 3! on the mRNA.
ii) Label the arrow with either the N (amino) termini of the protein being made or
the C (carboxyl) termini of the protein being made.
iii) As drawn, label the template strand for transcription.
iv) Box the three bases encoding the first amino acid of the protein being made.
v) Circle the part of the schematic where tRNAs would bind.
b) Give the anticodon used in the tRNA encoding Trp. Be sure to label the 5! and 3! ends.
c) Would a substitution within a codon for Trp always change the resulting protein
sequence? Explain your answer.
d) Would a substitution within a codon for Thr always change the resulting protein
sequence? Explain your answer.
Transcription and Translation
(4.2) Transcription, RNA processing, and translation in eukaryotes
Some important notes:
The processes of DNA replication, transcription, and translation are highly conserved in
prokaryotes and eukaryotes. An important difference, which is reflected in the
following problems, is the structure of eukaryotic genes.
Eukaryotic genes often have DNA that is transcribed but not translated. This DNA is
termed intervening sequences or introns. The DNA sequences that are represented by
the mRNA that is translated into protein are considered exons.
Diagnostic Question:
a) Shown below is a schematic of the production of a polypeptide. At the top is the
chromosomal arrangement found in the cell; a schematic of the polypeptide is shown
below it.
i) Label the process indicated by each arrow.
ii) Indicate on the diagram below where you would expect to find each of the
following components:
• Promoter
• Transcription terminator
• Start codon
• Stop codon
• Introns
• Exons
Gene 1
Gene 2
Chapter 3: Molecular Biology Problems
Answer to Diagnostic Question:
a) Shown below is a schematic of the production of a polypeptide. At the top is the
chromosomal arrangement found in the cell; a schematic of the polypeptide is shown
below it.
Gene 1
Gene 2
Promoter
Start codon
Promoter
Transcription terminator
Transcription
Splicing
Stop codon
Translation
Transcription
Splicing
Start codon
Translation
= introns
= exons for gene 1
= exons for gene 2
Transcription and Translation in Eukaryotes
(4.2.1) The alcohol dehydrogenase enzyme functions in the breakdown of alcohol in the
liver by converting it to acetylaldehyde. The reaction is outlined below:
ethanol
Ethanol ++ NAD
+
alcohol dehydrogenase
acetylaldehyde + NADH
The enzyme binds the cofactor NAD+ in order to carry out the oxidation of ethanol. The
utilization of NAD+ is crucial to the activity of the enzyme. The gene that encodes the
374-amino-acid enzyme is made up of 10 exons and 9 introns. Amino acid (aa) residues
encoded by exons 5 to 9 are involved in binding the NAD+ cofactor, while amino acid
residues encoded by exons 1, 2, 3, 4, and 10 are involved in catalysis. A schematic
diagram of the alcohol dehydrogenase gene is shown below.
338
aa
11
1
56
57
2
71
3
Exons
206 207
179
4
5
232
6
252 253
7
8
284 285
339 374
9
10
Introns
Note that codons can be interrupted by introns and later formed when the intron is
excised and two exons are joined together. As an example, see amino acid 11.
a) Using a format similar to the one given in the diagram above, draw the fully
processed mRNA of the alcohol dehydrogenase gene. Indicate the 5! and 3! ends of the
mRNA.
b) The DNA structure and partial codong-strand sequence of an exon/intron boundary
from a region in the gene are shown below. (Note: not all 148 nucleotides in the intron
are shown.)
Exon
Exon 5
5
Intron 5 (148 nt)
CAGGTAAGT
Exon 6 Exon 6
CAGGAA
i) Using the format given above, indicate the structure and partial sequence of
the mRNA at this exon/intron boundary from this region of the gene before and
after splicing of the mRNA.
Chapter 3: Molecular Biology Problems
ii) What are amino acids 206 and 207 in the alcohol dehydrogenase enzyme?
c) Certain individuals possess a defective alcohol dehydrogenase gene. This defective
gene produces a mutant enzyme. The defect is due to a mutation at the 5! splice site in
intron 5. The structure and partial sequence of this region in the DNA are shown below.
Exon
Exon
5 5
Intron 5 (148 nt)
CAGAAAAGT
Exon 6Exon 6
CAGGAA
i) Using a format similar to the one used in part (b), indicate the structure of the
entire spliced mRNA that encodes this mutant enzyme.
ii) What two possible effects can this mutation have on the structure of the
protein?
(4.2.2) The figures on the next two pages present a small gene that was found in the
eukaryotic fungus Neurospora crassa. This is intended as a real-world example of a
eukaryotic gene. Note that most genes in eukaryotes are substantially larger.
Transcription and Translation in Eukaryotes
5!-CGGTGAATAAATACGTCATGACGGTGCTGTCAGCATCATCGATAGGTAGGAGCGAACAAACAACCTAACATCGGATTGCA
1 +---------+---------+---------+---------+---------+---------+---------+--------3!-GCCACTTATTTATGCAGTACTGCCACGACAGTCGTAGTAGCTATCCATCCTCGCTTGTTTGTTGGATTGTAGCCTAACGT
GGACCGCGGGGCAGGATTGCTCCGGGCTGTTTCATGACTTGTCAGGTGGGATGACTTGGATGGAAAAGTAGAAGGTCATG
81 +---------+---------+---------+---------+---------+---------+---------+--------CCTGGCGCCCCGTCCTAACGAGGCCCGACAAAGTACTGAACAGTCCACCCTACTGAACCTACCTTTTCATCTTCCAGTAC
GGGTGGCCAACTTGGGCGAGAAAAGGTATATAAAGGTCTCTTGCTCCCATCAACTGCCTCAAAAGTAGGTATTCCAGCAG
161 +---------+---------+---------+---------+---------+---------+---------+--------CCCACCGGTTGAACCCGCTCTTTTCCATATATTTCCAGAGAACGAGGGTAGTTGACGGAGTTTTCATCCATAAGGTCGTC
ATCAGACAACCAAACAAACACACTTCATTCCCAAGACATCACTCACAAACAACCAACCTCTTCCAATCCAACCACAAACA
241 +---------+---------+---------+---------+---------+---------+---------+--------TAGTCTGTTGGTTTGTTTGTGTGAAGTAAGGGTTCTGTAGTGAGTGTTTGTTGGTTGGAGAAGGTTAGGTTGGTGTTTGT
AAAATCAGCCAATATGTCCGACTTCGAGAACAAGAACCCCAACAACGTCCTTGGCGGACACAAGGCCACCCTTCACAACC
321 +---------+---------+---------+---------+---------+---------+---------+--------TTTTAGTCGGTTATACAGGCTGAAGCTCTTGTTCTTGGGGTTGTTGCAGGAACCGCCTGTGTTCCGGTGGGAAGTGTTGG
CTAGTATGTATCCTCCTCAGAGCCTCCAGCTTCCGTCCCTCGTCGACATTTCCTTTTTTTTCATATTACATCCATCCAAG
401 +---------+---------+---------+---------+---------+---------+---------+--------GATCATACATAGGAGGAGTCTCGGAGGTCGAAGGCAGGGAGCAGCTGTAAAGGAAAAAAAAGTATAATGTAGGTAGGTTC
TCCCACAATCCATGACTAACCAGAAATATCACAGATGTTTCCGAGGAAGCCAAGGAGCACTCCAAGAAGGTGCTTGAAAA
481 +---------+---------+---------+---------+---------+---------+---------+--------AGGGTGTTAGGTACTGATTGGTCTTTATAGTGTCTACAAAGGCTCCTTCGGTTCCTCGTGAGGTTCTTCCACGAACTTTT
CGCCGGCGAGGCCTACGATGAGTCTTCTTCGGGCAAGACCACCACCGACGACGGCGACAAGAACCCCGGAAACGTTGCGG
561 +---------+---------+---------+---------+---------+---------+---------+--------GCGGCCGCTCCGGATGCTACTCAGAAGAAGCCCGTTCTGGTGGTGGCTGCTGCCGCTGTTCTTGGGGCCTTTGCAACGCC
GAGGATACAAGGCCACCCTCAACAACCCCAAAGTGTCCGACGAGGCCAAGGAGCACGCCAAGAAGAAGCTTGACGGCCTC
641 +---------+---------+---------+---------+---------+---------+---------+--------CTCCTATGTTCCGGTGGGAGTTGTTGGGGTTTCACAGGCTGCTCCGGTTCCTCGTGCGGTTCTTCTTCGAACTGCCGGAG
GAGTAAGCTCAGAGTTCACGAAAGAACCATTCGACGAGGGGAAGCACGGGGTTATCTCGTTCGAAACATGGGCCTGGTTA
721 +---------+---------+---------+---------+---------+---------+---------+--------CTCATTCGAGTCTCAAGTGCTTTCTTGGTAAGCTGCTCCCCTTCGTGCCCCAATAGAGCAAGCTTTGTACCCGGACCAAT
ATGCAAATGCATAATGGGGAGGATAATGAATCATGAGGTGTACGATATGGACGATATTGACGGATCTTAATTTGATGACA
801 +---------+---------+---------+---------+---------+---------+---------+--------TACGTTTACGTATTACCCCTCCTATTACTTAGTACTCCACATGCTATACCTGCTATAACTGCCTAGAATTAAACTACTGT
GTAATGAAATCACACCATAGT-3!
881 +---------+---------+ 901
CATTACTTTAGTGTGGTATCA-5!
Figure 1: Genomic DNA sequence of con-6 gene from Neurospora crassa. The sequence of
both strands (5! to 3! on top, 3! to 5! on bottom) is shown above with nucleotides
numbered 1 to 901. The dashed lines are interrupted every tenth nucleotide with a “+.”
Chapter 3: Molecular Biology Problems
GENOMIC DNA:
mRNA:
GENOMIC DNA:
mRNA:
GENOMIC DNA:
mRNA:
GENOMIC DNA:
mRNA:
GENOMIC DNA:
mRNA:
GENOMIC DNA:
mRNA:
GENOMIC DNA:
mRNA:
GENOMIC DNA:
mRNA:
GENOMIC DNA:
mRNA:
GENOMIC DNA:
mRNA:
GENOMIC DNA:
mRNA:
GENOMIC DNA:
mRNA:
GENOMIC DNA:
mRNA:
251 CAAACAAACACACTTCATTCCCAAGACATCACTCACAAACAACCAACCTC 300
|||||||||||||||||||||||||||||||||||||||||||||||||
1 @AAACAAACACACUUCAUUCCCAAGACAUCACUCACAAACAACCAACCUC 49
301 TTCCAATCCAACCACAAACAAAAATCAGCCAATATGTCCGACTTCGAGAA 350
||||||||||||||||||||||||||||||||||||||||||||||||||
50 UUCCAAUCCAACCACAAACAAAAAUCAGCCAAUAUGUCCGACUUCGAGAA 99
351 CAAGAACCCCAACAACGTCCTTGGCGGACACAAGGCCACCCTTCACAACC 400
||||||||||||||||||||||||||||||||||||||||||||||||||
100 CAAGAACCCCAACAACGUCCUUGGCGGACACAAGGCCACCCUUCACAACC 149
401 CTAGTATGTATCCTCCTCAGAGCCTCCAGCTTCCGTCCCTCGTCGACATT 450
|||
150 CUA............................................... 152
451 TCCTTTTTTTTCATATTACATCCATCCAAGTCCCACAATCCATGACTAAC 500
..................................................
501 CAGAAATATCACAGATGTTTCCGAGGAAGCCAAGGAGCACTCCAAGAAGG 550
||||||||||||||||||||||||||||||||||||
153 ..............AUGUUUCCGAGGAAGCCAAGGAGCACUCCAAGAAGG 188
551 TGCTTGAAAACGCCGGCGAGGCCTACGATGAGTCTTCTTCGGGCAAGACC 600
||||||||||||||||||||||||||||||||||||||||||||||||||
189 UGCUUGAAAACGCCGGCGAGGCCUACGAUGAGUCUUCUUCGGGCAAGACC 238
601 ACCACCGACGACGGCGACAAGAACCCCGGAAACGTTGCGGGAGGATACAA 650
||||||||||||||||||||||||||||||||||||||||||||||||||
240 ACCACCGACGACGGCGACAAGAACCCCGGAAACGUUGCGGGAGGAUACAA 288
651 GGCCACCCTCAACAACCCCAAAGTGTCCGACGAGGCCAAGGAGCACGCCA 700
||||||||||||||||||||||||||||||||||||||||||||||||||
289 GGCCACCCUCAACAACCCCAAAGUGUCCGACGAGGCCAAGGAGCACGCCA 338
701 AGAAGAAGCTTGACGGCCTCGAGTAAGCTCAGAGTTCACGAAAGAACCAT 750
||||||||||||||||||||||||||||||||||||||||||||||||||
339 AGAAGAAGCUUGACGGCCUCGAGUAAGCUCAGAGUUCACGAAAGAACCAU 388
751 TCGACGAGGGGAAGCACGGGGTTATCTCGTTCGAAACATGGGCCTGGTTA 800
||||||||||||||||||||||||||||||||||||||||||||||||||
389 UCGACGAGGGGAAGCACGGGGUUAUCUCGUUCGAAACAUGGGCCUGGUUA 438
801 ATGCAAATGCATAATGGGGAGGATAATGAATCATGAGGTGTACGATATGG 850
||||||||||||||||||||||||||||||||||||||||||||||||||
439 AUGCAAAUGCAUAAUGGGGAGGAUAAUGAAUCAUGAGGUGUACGAUAUGG 488
851 ACGATATTGACGGATCTTAATTTGATGACAGTAATGAAATCACACCATAG 900
|||||||||||||||||||||||||
489 ACGAUAUUGACGGAUCUUAAUUUGAAAAAAAAAAAAAAAAAAAAAAAAAA 538
Figure 2: Sequence alignment of con-6 genomic DNA and mRNA sequences. The top
line of each pair of sequences is the sequence of con-6 genomic DNA. The genomic DNA
nucleotides are numbered as in Figure 1. The bottom line is the sequence of a con-6
mRNA isolated from Neurospora crassa. The nucleotide numbers of the mRNA begin at
the 5! end with #1, and end with #539 at the 3! end. Vertical dashes indicate nucleotides
identical in both sequences (not base pairs!). Dots indicate nucleotides in the genomic
sequence that are not found in the mRNA sequence (@ represents 5! G-cap).
Transcription and Translation in Eukaryotes
You should use the information to make a map of the con-6 gene that follows the format
shown below.
Maps have the following format; the numbers correspond to the numbers on the DNA
strands. The map shown below is for illustration purposes only and does not
correspond to the gene we are studying.
Start codon
|
1
|
10
|
20
exon 1
|
30
|
40
intron 1
|
|
50
60
Stop codon
|
70
|
80
|
90
exon 2
|
100
|
110
|
120
|
Here is what this map shows (and a list of all the features that a map must contain); note
that all the map coordinates are approximate.
• The promoter is indicated by the small black rectangle at position 20.
• The start of transcription is indicated by a bent arrow; in this gene it is at position
22 (roughly).
• Exon 1 is indicated by a labeled box; it starts at 22 and ends at 40.
• Intron 1 is indicated by a labeled blank space; it starts at 41 and ends at 72.
• Exon 2 is indicated by a labeled box; it starts at 73 and ends at 110.
• The end of transcription is indicated by the end of the last exon; here it is at 110.
• The terminator is indicated by a small black rectangle around position 110.
• The start codon is indicated by the start of the hatched region in exon 1; it is at
position 35.
• The stop codon is indicated by the end of the hatched region in exon 2; it is at
position 90.
• The coding region, the region that encodes the protein, is indicated by the
hatched parts of the exons; it extends from 35 to 90. Note that it does not include
the intron.
a) Make your map of the con-6 gene based on Figures 1 and 2 on the preceding pages;
be sure to include all the features shown on the example. If there are more than two
exons and one intron, be sure to include them as well.
b) What are the first five and last five amino acids of the con-6 protein?
Chapter 3: Molecular Biology Problems
(4.2.3) Shown below is the sequence of a short fictitious eukaryotic gene. Both strands of
DNA are shown.
After exhaustive studies, you have determined the following:
•
Transcription in this organism always starts at the sixth nucleotide after the
TATAA. That is, given the following sequence, the first nucleotide of the mRNA
would be X: TATAAnnnnnX (where n can be any nucleotide).
•
The intron splice sites have been well defined. Introns always begin with
GUAUGU and end with CAG or UAG. That is, given the following sequences,
the bases in bold will be in the mature mRNA while the underlined nucleotides
will be spliced out as an intron:
XXXXGUAUGU(X)nCAGXXXXXX
XXXXGUAUGU(X)nUAGXXXXXX
1:
•
The intron-splicing machinery processes the RNA from 5! to 3!. It finds the first
GUAUGU sequence and the first subsequent (moving 5! to 3!) CAG or UAG and
removes the intron between them. (Note that these rules and sequences are
similar to those of real eukaryotic organisms.)
•
This organism adds poly(A) tails immediately after the sequence GAAUAAAU.
Poly(A) tails are usually about seven nucleotides long.
•
RNA is processed in this sequence: transcription, then splicing, then
polyadenylation and capping.
5!-GAAGCTAGAGGTCAATACCTGTATAAATGAAAAGGCGCTGGTATGTCCGAATAGCATGCA
---------+---------+---------+---------+---------+---------+
3!-CTTCGATCTCCAGTTATGGACATATTTACTTTTCCGCGACCATACAGGCTTATCGTACGT
60
61:
GAACATGCCTCTGTATGTATTACTGTAGCTTTAAGGTACTACGTATGTCCGTATGTAATA
---------+---------+---------+---------+---------+---------+
CTTGTACGGAGACATACATAATGACATCGAAATTCCATGATGCATACAGGCATACATTAT
121:
AATAACTGTACAGTAACTAATGATGGTTGACGATACCCTCGGAATAAATGCGCATACGTA-3!
---------+---------+---------+---------+---------+---------+ 180
TTATTGACATGTCATTGATTACTACCAACTGCTATGGGAGCCTTATTTACGCGTATGCAT-5!
120
a) The promoter sequence is TATAA. Why wouldn’t the sequence TATA (or even
TATATA, for that matter) work as a promoter sequence? (Hint: remember that a
promoter is not just a place for RNA polymerase to bind; a promoter must indicate
which direction RNA polymerase must read the DNA.)
Transcription and Translation in Eukaryotes
b) What is the sequence of the mature mRNA from this gene?
c) What is the sequence of the protein produced from this gene?
For the following mutations, describe the changes to the mRNA sequence (either list the
new mRNA sequence or just list that of the altered areas) and give the sequence of the
protein produced by the mutant gene. Base pairs are numbered and “C/G base pair”
means: C on the top strand, G on the bottom. (The base pairs in question are highlighted
in bold.) Consider each mutation separately.
d) T/A base pair 52 is changed to A/T.
e) G/C base pair 41 is changed to T/A.
f) T/A base pair 86 is changed to G/C.
g) T/A base pair 127 is changed to A/T.
h) A/T base pair 24 is deleted.
Chapter 3: Molecular Biology Problems
(4.2.4) This question is based on some experiments that were described by Chan, A.C. et
al. in the journal Science (volume 264, page 1599; 1994). Shown below is an internal
portion of the genomic DNA sequence of a gene which produces a protein kinase (we
will talk about these later in the course) essential for proper immune system function,
shown as double-stranded base-paired DNA:
1824
1840
1860
1880
|.
*
.
*
.
*
.
*
.
*
.
*
5!..CCCTACAAGGCAGGCGCGGGCAGAGGCAGGTGGGCGGTGTGGTGGGGAGGGGGATGA
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||
3!..GGGATGTTCCGTCCGCGCCCGTCTCCGTCCACCCGCCACACCACCCCTCCCCCTACT
1881
1900
1920
1940
|
.
*
.
*
.
*
.
*
.
*
.
*
GGAGGAGGACACTGGTCACTCACAGGTGTCTCTGCCCGGCTTGAGCAGAAGATGAAAGGG
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
CCTCCTCCTGTGACCAGTGAGTGTCCACAGAGACGGGCCGAACTCGTCTTCTACTTTCCC
1941
|
.
*
CCGGAGGTCATG...3!
||||||||||||
GGCCTCCAGTAC...5!
Starting with nucleotide 1824 and proceeding to the right, this region encodes the
following amino acid sequence:
N-....Pro-Tyr-Lys-Lys-Met-Lys-Gly-Pro-Glu-Val-Met-...C
a) The region contains a single intron. Using the sequence data above, locate the region
encoding the intron within the above genomic DNA sequence. Using the numbering
scheme used above, what are the first and last nucleotides of the DNA region that
encodes the intron?
b) Does the intron you have identified in part (a) follow the “GT.....AG rule” described
in most textbooks (the first two nucleotides of an intron are usually GU and the last two
are usually AG)? Which is more compelling evidence for the intron’s boundaries, the
“rule” or the protein sequence data?
Transcription and Translation in Eukaryotes
c) A mutant form of this gene is known; it carries the recessive phenotype of complete
immune deficiency. The mutation is a single-base-pair substitution in the region
encoding the intron. The protein sequence of the corresponding region of the mutant
protein is shown below with differences from wild-type underlined; the mutation
results in the insertion of three amino acids into the middle of the protein.
N-...Pro-Tyr-Lys-Leu-Glu-Gln-Lys-Met-Lys-Gly-Pro-Glu-Val-Met-...C
i) Using the numbering scheme used above, what are the first and last
nucleotides of the DNA region that encodes the intron in the mutant?
ii)Assuming that introns always end with AG and that the splicing machinery
proceeds from 5! to 3!, what is the mutation in these individuals (use the
numbering scheme used above)?
Chapter 3: Molecular Biology Problems
(C2) Computer Activity 2: Gene Explorer (GeneX). This computer simulation of
eukaryotic gene expression will help you to understand:
• More about transcription and translation
• Eukaryotic gene structure
• Exons and introns
• The effects of mutations on gene expression
Introduction:
Up until now, you have worked with genes of increasing complexity on paper. This can
be time-consuming and error-prone. We will now move on to examine the structure
and function of the genes using a computer simulation that takes care of the tedious
details.
This simulation allows you to explore gene expression in a hypothetical simplified
eukaryote. Eukaryotic genes have promoters and terminators for controlling
transcription as well as start and stop codons for controlling translation. Although
promoter and terminator sequences are different in different organisms, the genetic
code, including the start and stop codons, is identical in all organisms. The promoter
and terminator sequences used in the hypothetical organism simulated by the gene
explorer are:
a promoter:
a terminator:
5!-TATAA-3!
5!-GGGGG-3!
|||||
|||||
3!-ATATT-5!
3!-CCCCC-5!
Transcription begins with
the first base pair to the right
of this sequence and
continues to the right.
Transcription ends
with the first base
pair to the left of this
sequence.
Therefore, a gene would look like this:
5!-TATAAXXXXXXXXXXXXXXXXXXXXXXXXXXXXXGGGGG-3!
|||||||||||||||||||||||||||||||||||||||
3!-ATATTXXXXXXXXXXXXXXXXXXXXXXXXXXXXXCCCCC-5!
Where the region shown as X’s would be transcribed into pre-mRNA.
Transcription and Translation in Eukaryotes
In addition, eukaryotic genes have a few features that prokaryotic genes do not have.
These are:
• Transcription produces an mRNA called a pre-mRNA that is not yet ready for
translation.
• This pre-mRNA is then processed in several steps to produce the mature mRNA
ready for translation:
o The introns are removed and the exons are joined; this is called mRNA
splicing. This is controlled by splice signal sequences. In real organisms,
these sequences are not well known. In general, introns start with 5!-GU-3!
and end with 5!-AG-3!. In the hypothetical organism simulated by the
Gene Explorer, introns start with 5!-GUGCG-3! and end with
5!-CAAAG-3!.
o A modified G nucleotide is added to the 5! end of the mRNA; this is called
the “cap.” In the Gene Explorer, this is not shown.
o Many A’s are added to the 3! end of the mRNA; this is called the poly(A)
tail. In real organisms, as many as 400 A’s can be added at a specific signal
sequence; the Gene Explorer adds 13 A’s as a tail to the 3! end of any
mRNA. Note that these A’s do not correspond to T’s in the DNA.
In previous problems, you did the work of expressing a gene by hand. Now that you
are familiar with how these processes work, the Gene Explorer will do all the tedious
work of:
• Finding the promoter and terminator
• Reading the DNA sequence to produce the pre-mRNA
• Finding the splice sites
• Splicing and tailing the mRNA
• Finding the start codon
• Translating the mRNA
The Gene Explorer will then allow you to make specific mutations in a gene sequence,
and it will then calculate and display their effects on the mRNA and protein. You do
not have to deal with all the details listed above; the Gene Explorer will take care of it
all. Researchers use tools like this to analyze the genes they are studying.
Chapter 3: Molecular Biology Problems
Procedure
Part I: The Gene Explorer
1) Access the “Gene Explorer” from this site
http://intro.bio.umb.edu/MOOC/jsGX/JsGenex_C2.html.
2) You can also download a stand-alone version of Genex from
http://intro.bio.umb.edu/GX/
3) You will see something like this:
• The promoter is
shaded in green.
• The terminator is
shaded in pink.
DNA
strands
RNA
strands
The exons are
shaded in colors in
both the pre-mRNA
and mRNA.
protein
4) You can select a base in the DNA strand and the Gene Explorer will show the
corresponding bases in the mRNA and the corresponding amino acid in the protein, if
applicable. To select a base, here are some notes:
• You can select bases only in the top strand of DNA. If you click anywhere else on
the gene panel, nothing will happen.
• It can be tricky to select a particular base on the first click; here are some tips:
o You can use the right-arrow key to move one base to the right or the leftarrow key to move one base to the left.
o To hit a particular base, click on the space between it and the next base.
For example, to select 25, click on the space between 25 and 26.
• The number of the selected base is shown at the bottom right of the Gene
Explorer.
Transcription and Translation in Eukaryotes
An example is shown below. Here, base 63 has been selected.
This DNA base
was clicked on.
This is the corresponding base
in the other strand of DNA.
This is the corresponding
base in the pre-mRNA; this
lines up with the DNA.
This is the corresponding
base in the mature mRNA.
Exons are colored to match
the exons in the pre-mRNA.
This is the corresponding
amino acid in the protein;
this lines up with the
mature mRNA.
The protein sequence shown in
blue is the sequence of the most
recent previous version of the
protein.
This shows the number of
the selected base.
5) You can edit the DNA in several ways:
• To delete the selected base, use the “delete” or the “backspace” key.
• To replace the selected base with another base, type a lowercase letter (a, g, c, or t).
• To insert bases to the left of the selected base, type an uppercase letter (A, G, C, or T).
6) When you change the DNA sequence, the pre-mRNA, mature mRNA, and protein
sequences are automatically updated. The previous protein sequence, the sequence of the
protein before the latest change, is shown in blue for comparison purposes.
7) There are several useful buttons on the Gene Explorer:
• (Button that looks like a keyboard) – Click this to bring up some buttons for
entering and editing DNA – this works on tablets including the iPad.
• Reset DNA Sequence – Click this to reset the DNA to the starting gene sequence.
• Enter New DNA Sequence – Click this to enter a new DNA sequence.
• Evaluate Answer – this is not used
Chapter 3: Molecular Biology Problems
Terminator
Exon 1
Intron 1
Exon 2
Intron 2
Exon 3
Exon 2
Exon 3
Stop Codon
|
10
.
|
20
.
promoter or terminator
Map Key:
.
Exon 1
A map of this gene would be:
|
30
.
.
|
50
.
|
60
start of transcription
|
40
Intron 1
.
Exon 2
|
70
.
|
80
Intron 2
|
90
exon
.
.
|
100
Exon 3
.
.
|
120
coding region
|
110
Note that each exon is color-coded identically in the pre-mRNA and mature-mRNA; introns are not colored.
Start Codon
5!-CCGAGAUUGAUGCCUUUGUCGGAUGUCGAGCGAGGACCUUAAGAAGGUGUGAAAAAAAAAAAAAA-3!
N-MetProLeuSerAspValGluArgGlyPro-C
Poly(A) tail
Exon 1
mature-mRNA and Protein:
5!-CCGAGAUUGAUGCCUUGUGCGAUAAGGUGUGUCCCCCCCCAAAGUGUCGGAUGUCGAGUGCGCGUGCAAAAAAAAACAAAGGCGAGGACCUUAAGAAGGUGUGA-3!
pre-mRNA: Exon Intron
10
20
30
40
50
60
70
80
90
100
110
120
.
|
.
|
.
|
.
|
.
|
.
|
.
|
.
|
.
|
.
|
.
|
.
|
5!-CAAGGCTATAACCGAGATTGATGCCTTGTGCGATAAGGTGTGTCCCCCCCCAAAGTGTCGGATGTCGAGTGCGCGTGCAAAAAAAAACAAAGGCGAGGACCTTAAGAAGGTGTGAGGGGGCGCTCGAT-3!
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
3!-GTTCCGATATTGGCTCTAACTACGGAACACGCTATTCCACACAGGGGGGGGTTTCACAGCCTACAGCTCACGCGCACGTTTTTTTTTGTTTCCGCTCCTGGAATTCTTCCACACTCCCCCGCGAGCTA-5!
Promoter
Here is the sequence and map of the normal, starting, gene; you can use these for reference.
225
Legend
Gene maps like the one on the preceding page have the following format: the numbers
correspond to the numbers on the DNA strands. The map shown below is for
illustration purposes only and does not correspond to the gene we are studying.
Start codon
|
|
|
exon 1
|
|
1
10
20
30
40
intron 1
|
|
50
60
Stop codon
|
|
|
70
80
90
exon 2
|
100
|
|
|
110
120
130
Here is what this map shows (and a list of all the features that a map must contain); note
that all the map coordinates are approximate.
• The promoter is indicated by the small solid black rectangle at position 20.
• The start of transcription is indicated by a bent arrow; in this gene it is at position
22 (roughly).
• Exon 1 is indicated by a labeled box; it starts at 22 and ends at 40.
• The start codon is indicated by the start of the hatched region in exon 1; it is at
position 35.
• The coding region, the region that encodes the protein, is indicated by the
hatched parts of the exons; it extends from 35 to 40 and 73 to 90. Note that it does
not include the intron.
• Intron 1 is indicated by a labeled blank space; it starts at 41 and ends at 72.
• Exon 2 is indicated by a labeled box; it starts at 73 and ends at 110.
• The stop codon is indicated by the end of the hatched region in exon 2; it is at
position 90.
• The end of transcription is indicated by the end of the last exon; here it is at 110.
• The terminator is indicated by a small solid black rectangle around position 110.
Part II: A eukaryotic gene
1) Look at the Gene Explorer Display; it shows the unmodified gene.
2) Click on a base in the DNA and look at the parts of the other strands that are
highlighted; these correspond to the base you clicked on. You can click on any base you
like or use the arrow keys to move one base to the left or right. The number of the
currently selected base is shown at the bottom of the Gene Explorer window.
Use these features to explore the normal gene. For each of the following, give a DNA
nucleotide number that fits the description and give the name of that part of the gene
(intron, exon, untranslated region, coding region, etc.). If that type of DNA base is
impossible, put impossible. There may be more than one right answer for each.
Chapter 3: Molecular Biology Problems
a) A DNA base that corresponds to bases in the pre-mRNA, mature mRNA, and protein.
DNA base number
Part of gene
b) A DNA base that corresponds to bases in the mature mRNA, and protein but not in
the pre-mRNA.
DNA base number
Part of gene
c) A DNA base that corresponds to a base in the pre-mRNA, but not in the mature
mRNA.
DNA base number
Part of gene
d) A DNA base that corresponds to bases in the pre-mRNA, and the mature mRNA but
not in the protein.
DNA base number
Part of gene
3) There are several important things to notice about this gene.
a) Note that the poly(A) tail, the string of A’s at the 3’ end of the mature mRNA, does
not have any corresponding bases in the DNA or the pre-mRNA. How did they get
there?
b) A common misconception is that introns do not split inside codons. This is not true,
as you can see if you look at introns 1 and 2. Why doesn’t it matter if an intron occurs in
the middle of a codon?
c) Another common misconception is that an intron has to be a multiple of three
nucleotides long. This is also not true, as you can see if you measure the length of
introns 1 and 2 or if you try inserting some bases in an intron and see that it has no
effect. Why is it that introns can be any number of nucleotides long?
Transcription and Translation in Eukaryotes
228
|
10
.
|
20
.
|
30
.
|
40
.
Intron 1
|
50
.
|
60
.
Exon 2
|
70
.
|
80
Intron 2
.
|
90
5) Why are some regions insensitive to mutation while others are sensitive to mutation?
.
Exon 1
.
|
100
Exon 3
4) Draw your map of regions that are insensitive to this type of mutation on the figure below:
Type of mutation:
.
|
110
.
|
120
3) Make a map of the parts of the gene that can be changed without changing the protein sequence (other features of the
gene can change). Choose one type of mutation (insertion, deletion, or substitution) to use in your studies. Using the map
below, mark off the parts of the gene that can suffer a mutation without affecting the protein sequence.
2) Try making mutations in different parts of the normal gene; these can be insertions, deletions, or substitutions of one
DNA base. Be sure to click “Reset DNA Sequence” before making a new mutation; that way, you can see the effects of
one mutation at a time. Remember that the protein sequence corresponding to the current gene is shown in black; the
previous protein sequence is shown in blue for comparison purposes.
1) Click “Reset DNA Sequence.”
You will now make several mutations in the gene and explore their effects.
Part III: Mutations
229
|
10
.
|
10
.
|
20
|
20
Exon 1
.
.
|
30
|
30
.
.
|
40
|
40
.
.
Intron 1
|
50
|
50
.
.
|
60
|
60
.
.
Exon 2
|
70
|
70
.
.
|
80
|
80
Intron 2
.
.
|
90
|
90
.
.
|
100
|
100
Exon 3
.
.
|
110
|
110
.
.
|
120
|
120
10) Describe the differences between the original and mutant gene maps and explain how the mutant protein is longer
even though the mutant gene is shorter.
.
Mutant Gene Map:
.
Original Gene Map:
9) Using the line below, draw a map of the mutant gene. You can click on various bases in the DNA to help locate
important parts of the gene. Indicate the location of the mutation in the DNA sequence with an asterisk (*). How does the
structure of the mutant gene differ from the original gene?
Mutant Protein Sequence:
Original Protein Sequence: N-Met-Pro-Leu-Ser-Asp-Val-Glu-Arg-Gly-Pro-C
8) What is the amino acid sequence of the mutant protein? How does it differ from the original sequence? What kind of
mutation is this? Remember that the previous protein sequence is shown in blue; if you follow the directions above
exactly, it will show the original protein sequence.
7) Now do mutation 1: deleting the T/A base pair at position 26.
• Select base number 26; be sure that you have selected the right base by looking at the number in the lower right of
the screen.
GCCTGTGC
• Type the delete or backspace key once.
||||||||
• Look at the sequence of the DNA strands; it should look like this: CGGACACG
• If it does not look like this, go back to step 6 and try again. Note that base pair 26 is now missing.
6) Click “Reset DNA Sequence.”
230
|
10
.
|
10
.
|
20
|
20
Exon 1
.
.
|
30
|
30
.
.
|
40
|
40
.
.
Intron 1
|
50
|
50
.
.
|
60
|
60
.
.
Exon 2
|
70
|
70
.
.
|
80
|
80
Intron 2
15) Explain why the pattern of introns and exons is different in the mutant gene.
.
Mutant Gene Map:
.
Original Gene Map:
.
.
|
90
|
90
.
.
|
100
|
100
Exon 3
.
.
|
110
|
110
.
.
|
120
|
120
14) Using the line below, draw a map of the mutant gene. Indicate the location of the mutation in the DNA sequence with
an asterisk (*). How does the structure of the mutant gene differ from the original gene?
Mutant Protein Sequence:
13) What is the amino acid sequence of the mutant protein? How does it differ from the original sequence?
Original Protein Sequence: N-Met-Pro-Leu-Ser-Asp-Val-Glu-Arg-Gly-Pro-C
12) Now do mutation 2: changing the A/T base pair at position 51 to a T/A base pair.
• Select base number 51; be sure that you have selected the right base by looking at the number in the lower right of
CCCTAAGT
the screen.
||||||||
• Type “t” (be sure it is lowercase).
• Look at the sequence of the DNA strands; it should look like this: GGGATTCA
• Note that base pair 51 is now T in the top strand and A in the bottom strand.
• If it does not look like this, go back to step 11 and try again.
11) Click “Reset DNA Sequence.”
231
|
10
.
|
10
.
|
20
|
20
Exon 1
.
.
|
30
|
30
.
.
|
40
|
40
.
.
Intron 1
|
50
|
50
.
.
|
60
|
60
.
.
Exon 2
20) Explain why both the start and stop codons have now moved.
.
Mutant Gene Map:
.
Original Gene Map:
|
70
|
70
.
.
|
80
|
80
Intron 2
.
.
|
90
|
90
.
.
|
100
|
100
Exon 3
.
.
|
110
|
110
.
.
|
120
|
120
19) Using the line below, draw a map of the mutant gene. Indicate the location of the mutation in the DNA sequence with
an asterisk (*). How does the structure of the mutant gene differ from the original gene?
Mutant Protein Sequence:
18) What is the amino acid sequence of the mutant protein? How does it differ from the original sequence?
Original Protein Sequence: N-Met-Pro-Leu-Ser-Asp-Val-Glu-Arg-Gly-Pro-C
17) Now do mutation 3: changing the T/A base pair at position 21 to a G/C base pair.
• Select base number 21; be sure that you have selected the right base by looking at the number in the lower right of
TGAGGCCT
the screen.
||||||||
• Type “g” (be sure it is lowercase).
• Look at the sequence of the DNA strands; it should look like this: ACTCCGGA
• Note that base pair 21 is now G in the top strand and C in the bottom strand.
• If it does not look like this, go back to step 16 and try again.
16) Click “Reset DNA Sequence.”
232
|
10
.
|
10
.
|
20
|
20
.
.
|
30
|
30
.
.
|
40
|
40
.
.
Intron 1
|
50
|
50
.
.
|
60
|
60
.
.
Exon 2
|
70
|
70
.
.
|
80
|
80
Intron 2
23) Explain how the mutation caused the change you diagrammed in part (22).
.
Mutant Gene Map:
.
Exon 1
Draw a map of the resulting gene:
Mutant Protein Sequence:
Original Gene Map:
•
What did you change it to?
•
.
.
|
90
|
90
.
.
|
100
|
100
Exon 3
.
.
|
110
|
110
Original Protein Sequence: N-Met-Pro-Leu-Ser-Asp-Val-Glu-Arg-Gly-Pro-C
Which base pair did you change?
•
22) Invent a nonsense mutation of your own.
21) Click “Reset DNA Sequence.”
.
.
|
120
|
120
233
|
10
.
|
10
.
|
20
|
20
.
|
30
|
30
.
.
|
40
|
40
.
.
Intron 1
|
50
|
50
.
.
|
60
|
60
.
.
|
70
|
70
.
.
|
80
|
80
Intron 2
26) Explain how the mutation caused the change you diagrammed in part (25).
.
Mutant Gene Map:
.
.
Draw a map of the resulting gene:
•
Exon 2
Is the mutant protein sequence the same as the original?
Exon 1
How long is the mature mRNA in the mutant?
•
•
•
Original Gene Map:
Which base pair did you delete?
•
.
.
|
90
|
90
.
.
|
100
|
100
Exon 3
.
.
|
110
|
110
.
.
|
120
|
120
25) Invent a mutation where a deletion of one base in the DNA causes the mature mRNA to be longer. Note that the
normal mRNA [including the poly(A) tail] is about 65 nucleotides long; you can use the tick marks on the DNA strand to
estimate the length of the mature mRNA.
24) Click “Reset DNA Sequence.”
234
|
10
.
|
10
.
|
20
|
20
.
|
30
|
30
.
.
|
40
|
40
.
.
Intron 1
|
50
|
50
.
.
|
60
|
60
.
.
Exon 2
|
70
|
70
.
.
|
80
|
80
Intron 2
.
.
|
90
|
90
.
.
|
100
|
100
Exon 3
.
.
29) Explain how the mutation caused no mRNA to be made and how it caused no protein to be made.
.
Mutant Gene Map:
.
.
Draw a map of the resulting gene:
•
Exon 1
What did you change it to?
•
Original Gene Map:
Which base pair did you change?
•
|
110
|
110
.
.
|
120
|
120
28) Invent a mutation that alters only one DNA base (insertion, deletion, or substitution) that results in no mRNA or
protein being made at all.
27) Click “Reset DNA Sequence.”
235
|
10
.
|
10
.
|
20
|
20
.
.
|
30
|
30
.
.
|
40
|
40
.
.
|
50
|
50
.
.
|
60
|
60
.
.
Exon 2
|
70
|
70
.
.
|
80
|
80
Intron 2
32) Explain how the mutation caused the change you diagrammed in part (31).
.
Mutant Gene Map:
.
Intron 1
Draw a map of the resulting gene:
•
Exon 1
What was the change?
•
Original Gene Map:
Which base pair was changed?
•
.
.
What mutation led to this? (There may be more than one right answer; give only one.)
Note that the first six amino acids are the same as the normal protein (underlined).
|
90
|
90
N-MetProLeuSerAspValAspAlaArgAlaLysLysAsnLysGlyGluAspLeuLysLysVal-C
.
.
|
100
|
100
Exon 3
.
.
|
110
|
110
.
.
|
120
|
120
31) A single base mutation (one base inserted, deleted, or changed) in the starting gene results in the following protein
sequence:
30) Click “Reset DNA Sequence.”
This is a very challenging problem.
33) Here are some other mutations to try. For each one, explain how the mutation has
the effect described.
a) Make a mutation where one base is deleted that causes the protein to be longer.
b) Make a mutation where one base is inserted that causes the protein to be shorter.
c) Make a mutation that causes Exon 2 to be absent from the mature mRNA.
34) Design an entirely new gene that you have invented. Using the new Gene Explorer,
this gene should (you can make it more challenging if you like):
• Produce a protein of at least five amino acids (including the N-terminal Met).
• Contain at least one intron.
Tips
• Use the “Enter New DNA Sequence” button and delete the starting sequence
from the entry blank.
• Type in a promoter, a little DNA, and a terminator; be sure your RNA is made.
• Click on your gene and add the start codon, coding region, and stop codon; be
sure your protein is made. Type slowly so that the program can keep up.
• Similarly, add an intron in the coding region and be sure your gene works.
Chapter 3: Molecular Biology Problems
(5) CHALLENGE PROBLEMS
(5.1) DNA synthesis cannot begin de novo. It requires a free 3!-OH group. This free
OH is provided by an RNA primer. The RNA polymerase that makes this primer does
not require an end on which to build. DNA polymerase’s requirement for a primer has
an interesting effect on DNA replication. The final 3! end of the lagging strand cannot be
replicated, because there is no DNA left from which to make the RNA primer.
5
’
DNA synthesized
time =
before
1
5
’
time =
1
3
’
DNA synthesized
time = 1 and time =
between
2
3
’
5
’
time =
2
3
’
5
’
3
’
5
’
DNA synthesized
time = 2 and time =
between
3
Not enough room for primer and new
DNA
time =
3
3
’
5
’
If this problem were left uncorrected, linear chromosomes (which are long stretches of
DNA) would shrink with each successive replication. You might guess correctly that
this would have a detrimental effect on an organism. Therefore, organisms have
evolved ways to combat this problem.
Challenge Problems
a) Some organisms, such as bacteria and viruses, have circular, not linear, chromosomes.
Explain how having a circular chromosome could solve the shrinkage problem
explained above.
b) Another strategy that organisms take to combat shrinkage is to have linear
chromosomes with telomeres at either end. A telomere is simply a long stretch of
repeated nucleotides. For example, in yeast (S. cerevisiae), there is a telomere composed
of many (TGTGTGTG)n repeats present at the end of each chromosome, where n can
equal several hundred. A special enzyme called telomerase periodically extends the
length of this repeat sequence without requiring a template. How could having a
telomere solve the problem of shrinking chromosomes?
(5.2) Compare DNA and RNA.
a) List three key differences between DNA and RNA structures.
b) What molecular interaction allows base pairing to occur between two of the strands
of DNA in a double-stranded DNA helix?
c) DNA denaturation is the separation of the two strands of a DNA molecule. Consider
the two DNA sequences shown below. The symbol “|” indicates the molecular
interactions between the base pairs. Which sequence (i) or (ii) would you expect to
denature at a higher temperature? Briefly explain your reasoning.
(i)
(ii)
ATAGTATTC
|||||||||
TATCATAAG
GACGCGGTG
|||||||||
CTGCGCCAC
d) A DNA helix is formed from two DNA molecules with properly aligned base pairs.
Aside from base pairing, what other forces might contribute to the stability of the DNA
helix?
e) The usual base pair relationships are A-T and C-G. Which bases are purines and
which are pyrimidines? How might the DNA double helix be affected by the base pair
mismatch of A-G or C-T.
f) In a cell that is at physiological pH, what would be the overall charge (positive or
negative) of a double-stranded DNA molecule? Briefly justify your answer.
Chapter 3: Molecular Biology Problems
g) You carry out a DNA replication reaction using a single-stranded DNA template,
DNA polymerase, a primer, and the four deoxyribonucleoside triphosphates. You then
add the nucleotide form of AZT (azidothymidine) to the reaction mixture. The structure
of AZT is very similar to deoxythymidine except that in AZT, the 3!-hydroxyl (OH)
group on the deoxyribose ring has been replaced by an azido (N3) group. The
nucleotide form of AZT is shown on the next page.
O
Nucleotide Form of AZT
H C
3
O
O
P
O
O
P
O
H
N
O
O
O
P
O
N
5'
O
O
O
1'
4'
H
H
H
H
2'
3'
N3
H
•
What would you expect to happen to DNA replication when you add the AZT
nucleotide to the reaction mixture? Briefly explain your reasoning.
•
Why might AZT help individuals who have cancer or who are infected with HIV
(human immunodeficiency virus)?
h) Consider the structure of the base, 2,6-diaminopurine, shown below.
NH 2
N
N
H 2N
N
N
H
2,6-diaminopurine
With which of the normal pyrimidines (C or T) would 2,6-diaminopurine be able to base
pair? Draw the base pair and indicate the hydrogen bonds that would be formed. (You
do not need to draw the structure of the sugar or the phosphate groups.)
Challenge Problems
(5.3) You are studying translation in a bacterial species that has a single gene encoding
the tryptophan tRNA (the trnA-trp gene). The wild-type sequence of this gene is shown
below. The portion of the gene that encodes the anticodon of the tRNA is boxed.
5! GTACCTGCACTGCATGCCTAGCTAGCCCTAGCCAGCCTAGCTAGCTAGCACCAA3!
3! CATGGACGTGACGTACGGATCGATCGGGATCGGTCGGATCGATCGATCGTGGTT5!
a) Below is drawn the folded tryptophan-tRNA (produced from this trnA-trp gene). It is
shown base pairing with an mRNA containing the codon that this tryptophan-tRNA
recognizes. Fill in the three boxes on the tRNA with the correct nucleotide sequence of
its anticodon. Then fill in the three boxes on the mRNA with the correct nucleotide
sequence of the codon currently being read. Be sure to label the ends of the mRNA to
show directionality.
tryptophan
3!
’
5!
’
b) Which strand of the double-stranded trnA-trp gene is used as a template when the
tryptophan tRNA is transcribed, the upper strand or the lower strand? Remember that
tRNAs are transcribed directly from genes; there is no mRNA intermediate made during
the production of a tRNA from its DNA sequence.
Chapter 3: Molecular Biology Problems
Here are two different ways to show the genetic code:
U
C
A
G
U
UUU phe (F)
UUC phe (F)
UUA leu (L)
UUG leu (L)
CUU leu (L)
CUC leu (L)
CUA leu (L)
CUG leu (L)
AUU ile (I)
AUC ile (I)
AUA ile (I)
AUG met (M)
GUU val (V)
GUC val (V)
GUA val (V)
GUG val (V)
UCU
UCC
UCA
UCG
CCU
CCC
CCA
CCG
ACU
ACC
ACA
ACG
GCU
GCC
GCA
GCG
C
ser (S)
ser (S)
ser (S)
ser (S)
pro (P)
pro (P)
pro (P)
pro (P)
thr (T)
thr (T)
thr (T)
thr (T)
ala (A)
ala (A)
ala (A)
ala (A)
A
UAU tyr (Y)
UAC tyr (Y)
UAA STOP
UAG STOP
CAU his (H)
CAC his (H)
CAA gln (Q)
CAG gln (Q)
AAU asn (N)
AAC asn (N)
AAA lys (K)
AAG lys (K)
GAU asp (D)
GAC asp (D)
GAA glu (E)
GAG glu (E)
Challenge Problems
G
UGU cys (C)
UGC cys (C)
UGA STOP
UGG trp (W)
CGU arg (R)
CGC arg (R)
CGA arg (R)
CGG arg (R)
AGU ser (S)
AGC ser (S)
AGA arg (R)
AGG arg (R)
GGU gly (G)
GGC gly (G)
GGA gly (G)
GGG gly (G)
U
C
A
G
U
C
A
G
U
C
A
G
U
C
A
G
Chapter 3: Molecular Biology Problems